Three steps
Q = mc(Tf-Ti) to get 16 to zero for both water and aluminum
Q = mL to turn water to ice
Q = mc(Tf-Ti) to get zero to -8 for both water and aluminum.
Don't know specific heats of water, ice or aluminum offhand and I don't know latent heat of fusion for water)
0.50 L of water at 16°C is poured into an aluminum ice cube tray of mass 0.250 kg at the same temperature. How much energy must be removed from this system by the refrigerator to turn the water into ice at -8°C?
2 answers
Specific heat capacity for aluminum = 920 Jkg−1K−1 Specificheatcapacityforice=2100 Jkg−1K−1
Latent heat for water = 3.33 × 105 Jkg−1
Latent heat for water = 3.33 × 105 Jkg−1
2 answers